Question

A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T/3 second

Answer 1

The acceleration of the ball will be g(accelaration due to gravity). Initial velocity will be 0(initially rest position).
Assume that in T seconds body travels h meters
By applying equations of motion we get

s= ut +(1/2)gT²
Applying values

h = (1/2)gT² ------[1]

in T/3 sec,
h1 = (1/2)g(T/3)^2
h1 = (1/2)g(T²/9) ----[2]
From the equations,
[2] ÷ [1]
we get h1 =h/9 distance from point of release.
therefore distance from ground is h-h/9 =8h/9

So the ball will be at a height of h/9 from the top of tower or in other words 8h/9 distance from ground